計算幾何論文報告 Minimum local disk cover sets
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This document presents an algorithm for finding the minimum local disk cover sets for broadcasting in heterogeneous wireless ad hoc networks. It defines the problem and introduces the concept of a "skyline set" to represent the solution. It then proves several lemmas about the geometry of intersecting disks and uses mathematical induction to show that the number of arcs in any skyline set is upper bounded by 2n, where n is the number of disks. This allows the algorithm to run in O(n log n) time by using a divide-and-conquer approach to merge partial skyline sets.
計算幾何論文報告 Minimum local disk cover sets
- 1. Minimum Local Disk Cover Sets for Broadcasting in Heterogeneous Wireless Ad Hoc Networks 2007 International Conference on Parallel Processing Workshops (ICPPW 2007) Chih-Wei Yi, Fang-Chun Liu, Min-Te Sun 報告人 楊翔雲、李政遠
- 2. Definition
- 3. Definition
- 5. Skyline Set – for MLDC • Skyline is composed of arcs of disks. • Lemma 1: For any point a ∈ ∂B (ui, ri), the line segment oa ⊂ B (ui, ri). Proof. o ∈ B (ui, ri) and ||o − ui|| ≤ ri B (ui, ri) is concave and o, a ∈ B (ui, ri), the line segment oa ⊂ B (ui, ri).
- 6. Skyline Set – for MLDC • Corollary 2 A ray from o intersects the skyline at exactly one point • Proof. by contradiction. Assume a is farther from o than b. Since a is in the skyline, a is in ∂B (ui, ri) for some i. According to Lemma 1, we have oa ⊂ B (ui, ri). This implies b is inside of B (ui, ri), and b cannot be in the skyline.
- 7. Skyline Set – expr for MLDC • Define arc = (αi, uj, rj, αi+1), 0 < αi < 2π • Define skyline = (α0, us0, rs0, α1, us1, rs1, α2, ..., αn) AC divide to AB and BC
- 8. Skyline Set – expr for MLDC • Theorem 3 For a given local set V = {u0, u1, . . . , un}, its minimum local disk cover set is the skyline set for the corresponding local disk set {B(u0, r0), B(u1, r1), . . . , B(un, rn)}. • First, prove skyline set is a local disk cover set. 聯集結果相同。 • Second, prove if B (ui, ri) is in the skyline set, B (ui, ri) must be in any disk cover set. 任何在 skyline set 的 ui 會在任何解集合中(包括 最佳解)，意即 B (ui, ri) 存在自己的特別區域。
- 9. D&C algorithm – Skyline
- 10. D&C algorithm – Merge • 對齊一開始合併的角度，先根據 SL1 合併，之 後在合併 SL2，分別殘存的 arc list 再進行 merge
- 11. D&C algorithm – Merge-Case1 • Arcs (α, u, ru, β) and (α, v, rv, β) have no intersection. One arc is closer to o than the other, and the arc closer to o can not be in the new skyline. For instance. • 完全被包含
- 12. D&C algorithm – Merge-Case2 • Arcs (α, u, ru, β) and (α, v, rv, β) intersect at one point e. Let γ be the angle corresponding to the intersection point. Applying the principle used in Case 1, new skyline arcs can be decided from arcs (α, u, ru, γ) and (α, v, rv, γ), and arcs (γ, u, ru, β) and (γ, v, rv, β). • 恰好交一點，其中一部分的 arcs 隱入另外一個， 2 arcs 切分成 4 arcs
- 13. D&C algorithm – Merge-Case3 • Arcs (α, u, ru, β) and (α, v, rv, β) intersect at two points e, f. Let γ1 and γ2 with γ1 < γ2 be angles corresponding to intersection points. Applying the principle used in Case 1, new skyline arcs can be decided from arcs (α, u, ru, γ1) and (α, v, rv, γ1); arcs (γ1, u, ru, γ2) and (γ1, v, rv, γ2); and arcs (γ2, u, ru, β) and (γ2, v, rv, β). • 恰好交兩點，中間一段的 arcs 隱入另外一個，2 arcs 切分成 4 arcs
- 14. D&C algorithm – Analysis • Merge() time complexity: f(n) • f(n) is linear with respect to #arcs. According to the master theorem T(n) = O(n log n) • homogeneous networks < heterogeneous networks • heterogeneous networks is at least O(n ln n), O(n log n) algorithm is optimal time complexity for MLDS.
- 15. Geometry of Disks – in general • #arcs in a skyline can not be more than 2n • In fact, a disk may increase arcs more than two arcs. • 之前是依靠每一次加入的圓，最多貢獻 2 arcs。 來確保整體不超過 2n arcs。f (n) = O(n) • 如果從半徑由大到小依序 Merge()，則 Merge 過 程中整體的 arcs 不超過 2n 個。即使有一個圓貢 獻大於 2 個 arcs。
- 16. Geometry of Disks – in general • Lemma 5 In the skyline of {B1,B2, . . . , Bn}, if Bn contributes at least three arcs, then we can pick Bi, Bj, Bk from B1, B2, . . . , Bn−1 such that Bn contributes three arcs in the skyline of Bi, Bj, Bk, Bn • 等下用來證明的 Case 狀況
- 17. Geometry of Disks – in general • Lemma 6 Assume two circle ∂B1 and ∂B2 have two intersection points a and d. Let ac (and ab, respectively) be a diameters of B1 (and B2, respectively), and c (and b, respectively) is a point in arc cd (bd respectively). See Fig. If the angle ∠cab is obtuse, we have • 包含兩點 bc 的圓半徑 必然大於 min(r1, r2)
- 18. Geometry of Disks – in general • Lemma 7 Given an acute or right triangle, for each edge of the triangle, draw a circle with the edge as a chord whose center is outside the triangle and radius is larger than the circumradius. Then, three circles have no intersection. • 用來證明包含三頂點的 圓半徑小於三個邊外圓半徑， 會造成三個外圓沒有相交。
- 19. Geometry of Disks – in general • Lemma 8 The number of arcs in a skyline of n disks is at most 2n. • prove by mathematical induction Basis
- 20. Geometry of Disks – in general • Prove Lemma 8 cont. • Assume that as n = k, the skyline contains at most 2k arcs. If we can show that after a disk Bk+1 is added into the set, the number of arcs in the skyline increases at most by two, then the new skyline contains no more than 2 (k + 1) arcs prove by mathematical induction. • Assume Bk+1 can contribute at least 3 arcs. According to Lemma 5, without loss of generality, we may assume that Bk+1 contributes 3 arcs in the skyline of B1, B2, B3, Bk+1.
- 21. Geometry of Disks – Lemma 8 • Prove Lemma 8 cont. • Figure (b) one of intersections with B1, B2, B3 (1) Δabc is an acute or right triangle (2) Δabc is an obtuse triangle 21
- 22. Geometry of Disks – Lemma 8 • Prove Lemma 8 cont. • Case 1: Δabc is an acute or right triangle. Let rc be the circumradius of Δabc. Bk+1 contains a, b, c, we have rk+1 > rc. Bk+1 is the smallest among B0, B1, B2, . . .,Bk+1 rc < rk+1 ≤ r1, r2, r3 By Lemma 7, if r1, r2, r3 are larger than rc, then B1, B2, B3 have no intersection. • Impossible Case. 22
- 23. Geometry of Disks – Lemma 8 • Prove Lemma 8 cont. • Case 2: Δabc is an obtuse triangle • Let rc be the circumradius of Δabc. Bk+1 is the smallest among B0, B1, B2, . . .,Bk+1 rk+1 ≤ r1, r2, r3 By Lemma 6, ∠cab is obtuse, we have ||b − c|| > 2 min(r1, r2). Bk+1 contains Δabc, we have rk+1 ≥ ½|| b − c ||. • Impossible Case. 23
- 24. Geometry of Disks – Lemma 8 • Prove Lemma 8 cont. • Figure (c) one of intersections with B1, B2, B3 Case 3. Bk+1 cover exactly 3 intersection points. • Bk+1 must have 3 intersection points with B3. This is impossible to happen. • 兩圓最多交兩點！ 24
- 25. Geometry of Disks – Lemma 8 • Prove Lemma 8 cont. • Case 4. If Bk+1 cover exactly 4 intersection points. • Since Bk+1 is smaller than B1, B2, B3, arcs bc of B3 outside B1, B2, arc be of B1 outside B3, and arc ef of B2 outside B3 are larger than π. • ∠ bac > π/2 • By Lemma 6, Bk+1 contains Δabc, we have rk+1 ≥ ½|| b − c ||. • rk+1 > min(r1, r2) ⇒ Impossible Case. 25
- 26. Geometry of Disks – Lemma 8 • Assume Bk+1 can contribute at least 3 arcs. According to Lemma 5, without loss of generality, we may assume that Bk+1 contributes 3 arcs in the skyline of B1, B2, B3, Bk+1. • Case 1~4 are impossible. • The number of arcs in the skyline B1,B2, · · · ,Bk+1 is at most 2(k + 1). By mathematical induction, we conclude that the number of arcs in the skyline of n disks is upper bounded by 2n. 26
- 27. Conclusions • Merge() time complexity: f(n) • f(n) is linear with respect to #arcs. • According to the master theorem T(n) = O(n log n) • If the disks are added into the disk set follows the descending order of radius, #arcs = O(2n) and f(n) = O(n) which proved in Lemma 8.